Cho a,b,c thõa mãn a+b+c=0. Chứng tỏ rằng ab+bc+ca =<0 giúp minh với mấy bn ơi .....
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(a+b+c)2=a2+b2+c2+2ab+2ac+2bc=0
mà a2+b2+c2>=0 =>2(ab+bc+ca)<=0
=>ab+bc+ca<=0
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![](https://rs.olm.vn/images/avt/0.png?1311)
Từ AB=BC=CA
suy ra: AB+BC=2CA > CA
BC + AC = 2AB > AB
AB+AC = 2BC > BC
Suy ra A,B,C là 3 đỉnh của 1 tam giác vì nó thỏa mãn tính chất: Tổng 2 cạnh lớn hơn cạnh còn lại.
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\(\left(a-1\right)\left(b-1\right)\left(c-1\right)=\left(a-1\right)\left(bc-b-c+1\right)\)
\(=abc-\left(ab+bc+ca\right)+a+b+c-1\)
\(=abc-abc+1-1=0\) (đpcm)
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\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\Rightarrow\dfrac{a+b}{ab}=\dfrac{b+c}{bc}=\dfrac{c+a}{ca}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{c}+\dfrac{1}{a}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{c}\\\dfrac{1}{a}=\dfrac{1}{b}\end{matrix}\right.\) \(\Rightarrow a=b=c\)
\(\Rightarrow M=\dfrac{a^2+a^2+a^2}{a^2+a^2+a^2}=1\)
Cho a,b,c thõa mãn : a^2 + b^2 +c^2 - ab -bc- ca = 0. Tính: P = (a-b)^2020 + (b-c)^2021 + (c-a)^2022
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\(a^2+b^2+c^2-ab-bc-ac=0\)
\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\) (1)
Mà: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\)
Nên PT (1) \(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(a-c\right)^2=0\end{matrix}\right.\)
=> a = b = c
\(P=\left(a-b\right)^{2020}+\left(b-c\right)^{2021}+\left(c-a\right)^{2022}\)
\(=\left(a-a\right)^{2020}+\left(b-b\right)^{2021}+\left(c-c\right)^{2022}\)
= 0
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Ta chứng minh:\(\sqrt{a+bc}\ge a+\sqrt{bc}\)
\(\Leftrightarrow a+bc\ge a^2+bc+2a\sqrt{bc}\)
\(\Leftrightarrow a\ge a^2+2a\sqrt{bc}\)\(\Leftrightarrow a\ge a\left(a+2\sqrt{bc}\right)\Leftrightarrow1\ge a+2\sqrt{bc}\Leftrightarrow a+b+c\ge a+2\sqrt{bc}\)
\(\Leftrightarrow b+c-2\sqrt{bc}\ge0\Leftrightarrow\left(\sqrt{b}-\sqrt{c}\right)^2\ge0\)(luôn đúng)
\(\Leftrightarrow\sqrt{a+bc}\ge a+\sqrt{bc}\)
CMTT\(\sqrt{b+ca}\ge b+\sqrt{ca}\)
\(\sqrt{c+ab}\ge c+\sqrt{ab}\)
\(\Leftrightarrow\sqrt{a+bc}+\sqrt{b+ca}+\sqrt{c+ab}\ge a+b+c+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=1+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\)Vậy ......
(Dấu = xảy ra (=) a=b=c=1/3